physics_magnetism_and_moving_charges
 Magnetism and Moving Charges Magnetic Field:Magnetic field is defined by the force it exerts on a moving charged particle.It is electromagnetic field that exerts a force on a moving charge. Magnetic field can be defined as torque it produces on a magnetic dipole. The magnetic field lines1. The magnetic field lines of a magnet ( or a solenoid) form continuous closed loops.2. The tangent to the field line at a given point represents the direction of the net magnetic field Bat that point.3. The larger the number of field lines crossing per unit area, the strongeris the magnitude of the magnetic field B.4. The magnetic field lines do not intersect.5. Bar magnet may be thought of as a large number of circulating currents in analogy with a solenoid6. Magnetic sources are inherently dipole sources - you can't isolate North or South "monopoles. Gauss’s law for magnetism : The net magnetic flux through any closed surface is zero. Earth magnetic field arise due to electrical currents produced by convective motion of metallic fluids (consisting mostly of molten iron and nickel) in the outer core of the earth. This is known as the dynamo effect. Magnetisation ( M) : It is net magnetic moment per unit volume.M=mnetV$M=\frac{{m}_{net}}{V}$ magnetic dipole moment (m) associated with a current loop=m = N I A$=m\phantom{\rule{0.22em}{0ex}}=\phantom{\rule{0.22em}{0ex}}N\phantom{\rule{0.22em}{0ex}}I\phantom{\rule{0.22em}{0ex}}A$ Magnetic Potential EnergyUm=- m . B ${U}_{m}=-\stackrel{\to }{\phantom{\rule{0.22em}{0ex}}\mathbf{m}\phantom{\rule{0.22em}{0ex}}}\phantom{\rule{0.22em}{0ex}}.\stackrel{\to }{\phantom{\rule{0.22em}{0ex}}\mathbf{B}\phantom{\rule{0.22em}{0ex}}}$ Magnetic Intensity is a quantity used in describing magnetic phenomenon in terms of their magnetic fields. The strength of the magnetic field at a given point can be given in terms of vector quantity called as magnetic intensity (H). It is aB0=𝜇0H Total magnetic field =B=𝜇0(H+Mz) Mz=𝜒 HB=𝜇0(1+𝜒)H $\begin{array}{l}{B}_{0}={𝜇}_{0}H\phantom{\rule{0.22em}{0ex}}\\ Total\phantom{\rule{0.22em}{0ex}}magnetic\phantom{\rule{0.22em}{0ex}}field\phantom{\rule{0.22em}{0ex}}=B={𝜇}_{0}\left(H+{M}_{z}\right)\phantom{\rule{0.22em}{0ex}}\\ {M}_{z}=𝜒\phantom{\rule{0.22em}{0ex}}H\\ B={𝜇}_{0}\left(1+𝜒\right)H\\ \\ \end{array}$B=𝜇0𝜇r H$B={𝜇}_{0}{𝜇}_{r}\phantom{\rule{0.22em}{0ex}}H$ 𝜇=𝜇0𝜇r=𝜇0(1+𝜒)=$𝜇={𝜇}_{0}{𝜇}_{r}={𝜇}_{0}\left(1+𝜒\right)=$Magnetic permeability of substanceB0${B}_{0}$=magnetic field due to solenoid 𝜒 =$𝜒\phantom{\rule{0.22em}{0ex}}=$ Magnetic Susceptibility. It measures how the magnetic material responds to an external magnetic field.1+𝜒=𝜇r=relative permiability $1+𝜒={𝜇}_{r}=relative\phantom{\rule{0.22em}{0ex}}permiability\phantom{\rule{0.22em}{0ex}}$ of the substance. The magnetising current IM${I}_{M}$ is the additional current that needs to be passed through the windings of the solenoid in the absence of the core which would give a B value as in the presence of thecore. When magnetic fields pass through magnetic materials which themselves contribute internal magnetic fields, ambiguities can arise about what part of the field comes from the external currents and what comes from the material itself. So magnetic intensity is defined H. H=B𝜇0-M$H=\frac{B}{{𝜇}_{0}}-M$whereM=magnetisation of material.B=net magnetic field. Geographic Meridian: The vertical plane containing the longitude circle and the axis of rotation of the earth is called the geographic meridian Magnetic Axis: The line joining the magnetic north and magnetic south poles of the earth is called magnetic axis of earth. Magnetic Equator of earth: The plane perpendicular to the magnetic axis of the earth and passing thru the points where the magnetic needle is parallel to the earth's surface intersect the earth's spherical surface in to a circle. This circle is called magnetic equator of the earth. Magnetic meridian Magnetic meridian of a place as the vertical plane which passes through the imaginary line joining the magnetic north and the south poles. Angle of Declination: At any place the acute angle between the magnetic meridian and the geographical meridian is called angle of declination. It is the angle between the true geographic north and the north shown by a compass needle. Magnetic declination varies both from place to place and with the passage of time. Angle of Dip: The angle of dip at a place is the angle between the direction of earth magnetic field and the horizontal in the magnetic maridian at that place. Ferromagnetism : Ferromagnetic substances are those which gets strongly magnetised whenplaced in an external magnetic field.They have strong tendency to movefrom a region of weak magnetic field to strong magnetic field, i.e., they get strongly attracted to a magnet. The individual atoms (or ions or molecules) in a ferromagnetic material possess a dipole moment as in a paramagnetic material. They interact with one another in such a way that they spontaneously align themselves in a common direction over a macroscopic volume called domain. The ferromagnetic property depends on temperature. At high enough temperature, a ferromagnet becomes a paramagnet. The temperature of transition from ferromagnetic to paramagnetism is called the Curie temperature (Tc)$\left({T}_{c}\right)$.The susceptibility above the Curie temperature, i.e., in the paramagnetic phase is described by:𝜒=CT-Tc$𝜒=\frac{C}{T-{T}_{c}}$ here T>Tc$T>{T}_{c}$ Curie TemperatureWhen ferromagnetic substance is heated its magnetization decreases with temperature. At particular temperature it looses its magnetisation completely. The temperature at which domain structure is destroyed and the ferromagnetic substance looses its magnetism is called curie temperature. Diamagnetism : Diamagnetic substances are those which have tendency to move from strongerto the weaker part of the external magnetic field. In other words, unlike the way a magnetattracts metals like iron, it would repel a diamagnetic substance. Magnetic Susceptibility is negative for Diamagnetic material. The most exotic diamagnetic materials are superconductors. These are metals, cooled to very low temperatures which exhibits both perfect conductivity and perfect diamagnetism. In superconductors 𝜒=-1 and 𝜇r=0 $𝜒=-1\phantom{\rule{0.22em}{0ex}}and\phantom{\rule{0.22em}{0ex}}{𝜇}_{r}=0\phantom{\rule{0.22em}{0ex}}$. A superconductor repels a magenet and is repelled by a magnet. The phenomenon of perfect diamagnetism in superconductor is called Meissner effect. Diamagnetic substances are the ones in which resultant magnetic moment in an atom is zero. Paramagnetic Material: Paramagnetic Material are those which gets weakly magnetized when placed in external magnetic field. Magnetic Susceptibility is small and positive for paramagnetic material. Magnetisation of a paramagnetic material is inversely proportional to the absolute temperature T.M=C B0T $M=C\phantom{\rule{0.22em}{0ex}}\frac{{B}_{0}}{T}\phantom{\rule{1.54em}{0ex}}$where C=Curie constant and T is absolute temperature.$C=Curie\phantom{\rule{0.22em}{0ex}}constant\phantom{\rule{0.22em}{0ex}}and\phantom{\rule{0.22em}{0ex}}T\phantom{\rule{0.22em}{0ex}}is\phantom{\rule{0.22em}{0ex}}absolute\phantom{\rule{0.22em}{0ex}}temperature.$𝜒=C 𝜇0T $𝜒=C\phantom{\rule{0.22em}{0ex}}\frac{{𝜇}_{0}}{T}\phantom{\rule{0.66em}{0ex}}$
 Questions and Answers Q.5.3A short bar magnet placed with its axis at 30°${30}^{°}$ with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5×10-2 J$4.5×{10}^{-2}\phantom{\rule{0.22em}{0ex}}J$. What is the magnitude of magnetic moment of the magnet?A.5.3Torque=𝜏=m B sin𝜃 $Torque=𝜏=m\phantom{\rule{0.22em}{0ex}}B\phantom{\rule{0.22em}{0ex}}sin𝜃\phantom{\rule{0.22em}{0ex}}$⟹4.5×10-2=m 0.25 sin30 $⟹4.5×{10}^{-2}=m\phantom{\rule{0.22em}{0ex}}0.25\phantom{\rule{0.22em}{0ex}}sin30\phantom{\rule{0.22em}{0ex}}$⟹m =4.5×10-2×20.25$⟹m\phantom{\rule{0.22em}{0ex}}=\frac{4.5×{10}^{-2}×2}{0.25}$⟹m =4.5×10-2×20.25$⟹m\phantom{\rule{0.22em}{0ex}}=\frac{4.5×{10}^{-2}×2}{0.25}$⟹m =0.36 Am2$⟹m\phantom{\rule{0.22em}{0ex}}=0.36\phantom{\rule{0.22em}{0ex}}A{m}^{2}$ Q.5.4A short bar magnet of magnetic moment m = 0.32 JT-1$J{T}^{-1}$ is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?A.5.4(a) Stable equilibrium: when angle between "short bar magnet" and "magnetic field" is 0 degree then it is in stable equilibriumMagnetic Potential EnergyUm=- m . B ${U}_{m}=-\stackrel{\to }{\phantom{\rule{0.22em}{0ex}}\mathbf{m}\phantom{\rule{0.22em}{0ex}}}\phantom{\rule{0.22em}{0ex}}.\stackrel{\to }{\phantom{\rule{0.22em}{0ex}}\mathbf{B}\phantom{\rule{0.22em}{0ex}}}$Um=-0.32×0.15 cos0=-0.048 J${U}_{m}=-0.32×0.15\phantom{\rule{0.22em}{0ex}}cos0=-0.048\phantom{\rule{0.22em}{0ex}}J$ (a) Unstable equilibrium: when angle between "short bar magnet" and "magnetic field" is 180 degree then it is in stable equilibriumMagnetic Potential EnergyUm=-0.32×0.15 cos180=0.048 J${U}_{m}=-0.32×0.15\phantom{\rule{0.22em}{0ex}}cos180=0.048\phantom{\rule{0.22em}{0ex}}J$ Q.5.19A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction east to west. The earth’s magnetic field at the place is 0.39 G, and the angle of dip is 35 degree. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable? A.5.19 Bearth=0.39 G${B}_{earth}=0.39\phantom{\rule{0.22em}{0ex}}G$magnetic field due to wireaB= 4×𝜇0 I2 𝜋 RB=4×4𝜋⨯10-7⨯12𝜋⨯0.04$\begin{array}{l}B=\phantom{\rule{0.22em}{0ex}}4×\frac{{𝜇}_{0}\phantom{\rule{0.22em}{0ex}}I}{2\phantom{\rule{0.22em}{0ex}}𝜋\phantom{\rule{0.22em}{0ex}}R}\\ B=\frac{4×4𝜋⨯{10}^{-7}⨯1}{2𝜋⨯0.04}\end{array}$ B=2⨯10-5 T =0.2 G$B=2⨯{10}^{-5}\phantom{\rule{0.22em}{0ex}}T\phantom{\rule{0.22em}{0ex}}=0.2\phantom{\rule{0.22em}{0ex}}G$westeastCurrent I=1.0 A35 degreenorthsouthB=0.2 G$B=0.2\phantom{\rule{0.22em}{0ex}}G$Bh=-0.2+0.39 cos35=0.12 G${B}_{h}=-0.2+0.39\phantom{\rule{0.22em}{0ex}}cos35=0.12\phantom{\rule{0.22em}{0ex}}G$Bv=0.39 sin35=0.224 G${B}_{v}=0.39\phantom{\rule{0.22em}{0ex}}sin35=0.224\phantom{\rule{0.22em}{0ex}}G$Bnet=(0.12)2+(0.224)2=0.254 G${B}_{net}=\left(0.12{\right)}^{2}+\left(0.224{\right)}^{2}=0.254\phantom{\rule{0.22em}{0ex}}G$Below the cable Bearth=0.39 G${B}_{earth}=0.39\phantom{\rule{0.22em}{0ex}}G$35 degreenorthsouthB=0.2 G$B=0.2\phantom{\rule{0.22em}{0ex}}G$Bh=0.2+0.39 cos35=0.52 G${B}_{h}=0.2+0.39\phantom{\rule{0.22em}{0ex}}cos35=0.52\phantom{\rule{0.22em}{0ex}}G$Bv=0.39 sin35=0.224 G${B}_{v}=0.39\phantom{\rule{0.22em}{0ex}}sin35=0.224\phantom{\rule{0.22em}{0ex}}G$Bnet=(0.52)2+(0.224)2=0.57 G${B}_{net}=\left(0.52{\right)}^{2}+\left(0.224{\right)}^{2}=0.57\phantom{\rule{0.22em}{0ex}}G$Above the cable Q.5.20 A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45°${45}^{°}$ with the magnetic meridian.When the current in the coil is 0.35 A, the needle points west to east.(a) Determine the horizontal component of the earth’s magnetic fieldat the location.(b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of 90°${90}^{°}$ in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero. A.5.20 Bearth${B}_{earth}$Bcoil${B}_{coil}$45 degreeMagnetic field at the center of the loopaBcoil = 𝜇0 N I2 RBcoil=4𝜋⨯10-7⨯30⨯0.352⨯0.12=549.5⨯10-7 Bcoil sin45=Bearth Bearth=388⨯10-7=0.388⨯10-4 T=0.388 G $\begin{array}{l}{B}_{coil}\phantom{\rule{0.22em}{0ex}}=\phantom{\rule{0.22em}{0ex}}\frac{{𝜇}_{0}\phantom{\rule{0.22em}{0ex}}N\phantom{\rule{0.22em}{0ex}}I}{2\phantom{\rule{0.22em}{0ex}}R}\\ {B}_{coil}=\frac{4𝜋⨯{10}^{-7}⨯30⨯0.35}{2⨯0.12}=549.5⨯{10}^{-7}\\ \\ {B}_{coil}\phantom{\rule{0.22em}{0ex}}sin45={B}_{earth}\\ \\ {B}_{earth}=388⨯{10}^{-7}=0.388⨯{10}^{-4}\phantom{\rule{0.22em}{0ex}}T=0.388\phantom{\rule{0.22em}{0ex}}G\\ \end{array}$ eastwest 5.21 A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60°${60}^{°}$, and one of the fields has a magnitude of 1.2 × 10-2${10}^{-2}$ T. If the dipole comes to stable equilibrium at an angle of 15°${15}^{°}$ with this field, what is the magnitude of the other field?A.5.21: Potential enerygy is minimum when 𝜃$𝜃$ is zero. So direction of resultant magnetic field must be in direction of angle of 15°${15}^{°}$ 15 degree60 degreeB1=1.2⨯10-2 T${B}_{1}=1.2⨯{10}^{-2}\phantom{\rule{0.22em}{0ex}}T$B2${B}_{2}$aB2sin60=B1sin15⟹B2sin60=1.2⨯10-2 sin15⟹B2=1.2⨯10-2 sin15sin45 ⟹B2=4.4⨯10-3 T $\begin{array}{l}{B}_{2}sin60={B}_{1}sin15\\ ⟹{B}_{2}sin60=1.2⨯{10}^{-2}\phantom{\rule{0.22em}{0ex}}sin15\\ ⟹{B}_{2}=\frac{1.2⨯{10}^{-2}\phantom{\rule{0.22em}{0ex}}sin15}{sin45}\\ \\ ⟹{B}_{2}=4.4⨯{10}^{-3}\phantom{\rule{0.22em}{0ex}}T\\ \end{array}$45 degree Q.5.22. A mono energetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.4 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm (me=9⨯10-31 kg${m}_{e}=9⨯{10}^{-31}\phantom{\rule{0.22em}{0ex}}kg$). A.5.22amass of electron=me=9⨯10-31 kgcharge of electron=qe=-1.6⨯10-19 C Let velocity of electron=vDistance travelled by beam=30 cm=0.3 mMagnetic field=B=0.4 G=0.4⨯10-4 T Kinetic energy of electron=12mv2=12⨯9⨯10-31 v2 Energy in electron=18 keV=18000⨯1.6⨯10-19 Joule So, 12⨯9⨯10-31 v2 =18000⨯1.6⨯10-19 ⟹v2=6400⨯1012⟹v=8⨯107 ms-1 Magnetic force=FB=qvBqvB=mv2r⟹r=mvqeB=9⨯10-31⨯8⨯107 1.6⨯10-19 ⨯0.4⨯10-4 ⟹r=11.3 m $\begin{array}{l}mass\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}electron={m}_{e}=9⨯{10}^{-31}\phantom{\rule{0.22em}{0ex}}kg\\ charge\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}electron={q}_{e}=-1.6⨯{10}^{-19}\phantom{\rule{0.22em}{0ex}}C\phantom{\rule{0.22em}{0ex}}\\ Let\phantom{\rule{0.22em}{0ex}}velocity\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}electron=v\\ Distance\phantom{\rule{0.22em}{0ex}}travelled\phantom{\rule{0.22em}{0ex}}by\phantom{\rule{0.22em}{0ex}}beam=30\phantom{\rule{0.22em}{0ex}}cm=0.3\phantom{\rule{0.22em}{0ex}}m\\ Magnetic\phantom{\rule{0.22em}{0ex}}field=B=0.4\phantom{\rule{0.22em}{0ex}}G=0.4⨯{10}^{-4}\phantom{\rule{0.22em}{0ex}}T\phantom{\rule{0.22em}{0ex}}\\ Kinetic\phantom{\rule{0.22em}{0ex}}energy\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}electron=\frac{1}{2}m{v}^{2}=\frac{1}{2}⨯9⨯{10}^{-31}\phantom{\rule{0.22em}{0ex}}{v}^{2}\phantom{\rule{0.22em}{0ex}}\\ \\ Energy\phantom{\rule{0.22em}{0ex}}in\phantom{\rule{0.22em}{0ex}}electron=18\phantom{\rule{0.22em}{0ex}}keV=18000⨯1.6⨯{10}^{-19}\phantom{\rule{0.22em}{0ex}}Joule\phantom{\rule{0.22em}{0ex}}\\ \\ So,\phantom{\rule{0.22em}{0ex}}\frac{1}{2}⨯9⨯{10}^{-31}\phantom{\rule{0.22em}{0ex}}{v}^{2}\phantom{\rule{0.22em}{0ex}}=18000⨯1.6⨯{10}^{-19}\phantom{\rule{0.22em}{0ex}}\\ \\ ⟹{v}^{2}=6400⨯{10}^{12}\\ ⟹v=8⨯{10}^{7}\phantom{\rule{0.22em}{0ex}}m{s}^{-1}\phantom{\rule{0.22em}{0ex}}\\ \\ Magnetic\phantom{\rule{0.22em}{0ex}}force={F}_{B}=qvB\\ qvB=\frac{m{v}^{2}}{r}\\ ⟹r=\frac{mv}{{q}_{e}B}=\frac{9⨯{10}^{-31}⨯8⨯{10}^{7}\phantom{\rule{0.44em}{0ex}}}{1.6⨯{10}^{-19}\phantom{\rule{0.22em}{0ex}}⨯0.4⨯{10}^{-4}\phantom{\rule{0.22em}{0ex}}}\\ \\ ⟹r=11.3\phantom{\rule{0.22em}{0ex}}m\phantom{\rule{0.22em}{0ex}}\\ \end{array}$ 0.3m0.3m11.3mACBD aarc(AC)=0.3mSo, ∠ABC=0.311.3 radian=∠ABC=0.02655 radian=1.522°BC=11.3 cos1.522°=11.296 m⟹AC=11.3-11.296⟹AC=0.004m=4 mm So deflection = 4 mm $\begin{array}{l}arc\left(AC\right)=0.3m\\ So,\phantom{\rule{0.22em}{0ex}}\angle ABC=\frac{0.3}{11.3}\phantom{\rule{0.22em}{0ex}}radian=\\ \angle ABC=0.02655\phantom{\rule{0.22em}{0ex}}radian=1.522°\\ BC=11.3\phantom{\rule{0.22em}{0ex}}cos{1.522}^{°}=11.296\phantom{\rule{0.22em}{0ex}}m\\ ⟹AC=11.3-11.296\\ ⟹AC=0.004m=4\phantom{\rule{0.22em}{0ex}}mm\\ \\ So\phantom{\rule{0.22em}{0ex}}deflection\phantom{\rule{0.22em}{0ex}}=\phantom{\rule{0.22em}{0ex}}4\phantom{\rule{0.22em}{0ex}}mm\\ \end{array}$ Q.5.23A sample of paramagnetic salt contains 2.0⨯1024$2.0⨯{10}^{24}$ atomic dipoles each of dipole moment1.5⨯10-23 JT-1$1.5⨯{10}^{-23}\phantom{\rule{0.22em}{0ex}}J{T}^{-1}$. The sample is placed under a homogeneous magnetic field of 0.64 T, and cooled to a temperature of 4.2 K. The degree of magnetic saturation achieved is equal to 15%.What is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K? (Assume Curie’s law) A.5.23 aNumber of dipoles=2.0⨯1024 Magnitude of each dipole moment=m=1.5⨯10-23 JT-1 So total dipole moment=2.0⨯1024 ⨯1.5⨯10-23 =30 JT-1Magnetic saturation=15%So effective dipole moment=15100⨯30=4.5JT-1 $\begin{array}{l}Number\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}dipoles=2.0⨯{10}^{24}\phantom{\rule{0.22em}{0ex}}\\ Magnitude\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}each\phantom{\rule{0.22em}{0ex}}dipole\phantom{\rule{0.22em}{0ex}}moment=m=1.5⨯{10}^{-23}\phantom{\rule{0.22em}{0ex}}J{T}^{-1}\phantom{\rule{0.22em}{0ex}}\\ So\phantom{\rule{0.22em}{0ex}}total\phantom{\rule{0.22em}{0ex}}dipole\phantom{\rule{0.22em}{0ex}}moment=2.0⨯{10}^{24}\phantom{\rule{0.22em}{0ex}}⨯1.5⨯{10}^{-23}\phantom{\rule{0.22em}{0ex}}=30\phantom{\rule{0.22em}{0ex}}J{T}^{-1}\\ Magnetic\phantom{\rule{0.22em}{0ex}}saturation=15%\\ So\phantom{\rule{0.22em}{0ex}}effective\phantom{\rule{0.22em}{0ex}}dipole\phantom{\rule{0.22em}{0ex}}moment=\frac{15}{100}⨯30=4.5J{T}^{-1}\\ \\ \\ \\ \end{array}$Magnetisation of a paramagnetic material is inversely proportional to the absolute temperature T.aM=C B0T $\begin{array}{l}M=C\phantom{\rule{0.22em}{0ex}}\frac{{B}_{0}}{T}\phantom{\rule{0.66em}{0ex}}\\ \phantom{\rule{0.22em}{0ex}}\end{array}$where C=Curie constant and T is absolute temperature.aSo here MTB is constant M1T1B1=M2T2B2⟹4.5⨯4.20.64=M2⨯2.80.98 ⟹M2=10.336 JT-1 $\begin{array}{l}So\phantom{\rule{0.22em}{0ex}}here\phantom{\rule{0.22em}{0ex}}\frac{MT}{B}\phantom{\rule{0.22em}{0ex}}is\phantom{\rule{0.22em}{0ex}}constant\\ \\ \frac{{M}_{1}{T}_{1}}{{B}_{1}}=\frac{{M}_{2}{T}_{2}}{{B}_{2}}\\ ⟹\frac{4.5⨯4.2}{0.64}=\frac{{M}_{2}⨯2.8}{0.98}\\ \\ ⟹{M}_{2}=10.336\phantom{\rule{0.22em}{0ex}}J{T}^{-1}\\ \end{array}$ Q.5.24A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field B in the core for a magnetising current of 1.2 A? A.5.24aB0=𝜇0 n I n=N2𝜋R =number of turns per unit length B0=𝜇0NI2𝜋R Magnetic Intensity=H=B0𝜇0=3500⨯1.22𝜋(0.15)=14000𝜋 B=𝜇0 𝜇r H=4𝜋⨯10-7⨯800⨯14000𝜋=4.48 T $\begin{array}{l}{B}_{0}={𝜇}_{0}\phantom{\rule{0.22em}{0ex}}n\phantom{\rule{0.22em}{0ex}}I\phantom{\rule{0.22em}{0ex}}\\ n=\frac{N}{2𝜋R}\phantom{\rule{0.22em}{0ex}}=number\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}turns\phantom{\rule{0.22em}{0ex}}per\phantom{\rule{0.22em}{0ex}}unit\phantom{\rule{0.22em}{0ex}}length\phantom{\rule{0.22em}{0ex}}\\ \\ {B}_{0}=\frac{{𝜇}_{0}NI}{2𝜋R}\\ \\ \\ \\ Magnetic\phantom{\rule{0.22em}{0ex}}Intensity=H=\frac{{B}_{0}}{{𝜇}_{0}}=\frac{3500⨯1.2}{2𝜋\left(0.15\right)}=\frac{14000}{𝜋}\\ \\ B={𝜇}_{0}\phantom{\rule{0.22em}{0ex}}{𝜇}_{r}\phantom{\rule{0.22em}{0ex}}H=4𝜋⨯{10}^{-7}⨯800⨯\frac{14000}{𝜋}=4.48\phantom{\rule{0.22em}{0ex}}T\phantom{\rule{0.22em}{0ex}}\\ \\ \end{array}$
 Questions and Answer Q.6.10A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth’s magnetic field at the location has a magnitude of 5⨯10-5 T $5⨯{10}^{-5}\phantom{\rule{0.22em}{0ex}}T\phantom{\rule{0.22em}{0ex}}$ and the dip angle is 30°.A.6.10aSpeed of plane = 1800km/hr=500 m/sec span of the wing=25 m angle of dip=30° B=5⨯10-5 sin30°=2.5⨯10-5 T ℇ=B l v =2.5⨯10-5 ⨯0.25⨯500 ℇ=3.125⨯10-3 Volt $\begin{array}{l}Speed\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}plane\phantom{\rule{0.22em}{0ex}}=\phantom{\rule{0.22em}{0ex}}1800km/hr=500\phantom{\rule{0.22em}{0ex}}m/sec\\ \\ span\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}the\phantom{\rule{0.22em}{0ex}}wing=25\phantom{\rule{0.22em}{0ex}}m\\ \\ angle\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}dip={30}^{°}\\ \\ B=5⨯{10}^{-5}\phantom{\rule{0.22em}{0ex}}sin{30}^{°}=2.5⨯{10}^{-5}\phantom{\rule{0.22em}{0ex}}T\\ \\ ℇ=B\phantom{\rule{0.22em}{0ex}}l\phantom{\rule{0.22em}{0ex}}v\phantom{\rule{0.22em}{0ex}}=2.5⨯{10}^{-5}\phantom{\rule{0.22em}{0ex}}⨯0.25⨯500\\ \\ ℇ=3.125⨯{10}^{-3}\phantom{\rule{0.22em}{0ex}}Volt\\ \end{array}$ Q.6.12A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm sec-1$8\phantom{\rule{0.22em}{0ex}}cm\phantom{\rule{0.22em}{0ex}}se{c}^{-1}$ in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10-3Tcm-1${10}^{-3}Tc{m}^{-1}$ along the negative x-direction (that is it increases by 10-3Tcm-1${10}^{-3}Tc{m}^{-1}$ as one moves in the negative x-direction), and it is decreasing in time at the rate of 10-3Tsec-1${10}^{-3}Tse{c}^{-1}$ . Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mΩ.A.6.12aHere B is changing with time (t) and with x . L=side of the square dBdx=-10-3 T cm-1=-0.1 Tm-1 dBdx=-10-3 T sec-1 dxdt=8 cm sec-1=0.08 m sec-1 𝜙B=B.A Emf due to changing of magnetic field with time ℇ1=-d(BA)dt=-AdBdt=-(0.12⨯0.12)⨯(-10-3)=0.0144⨯10-3 =1.44⨯10-5Volt $\begin{array}{l}Here\phantom{\rule{0.22em}{0ex}}B\phantom{\rule{0.22em}{0ex}}is\phantom{\rule{0.22em}{0ex}}changing\phantom{\rule{0.22em}{0ex}}with\phantom{\rule{0.22em}{0ex}}time\phantom{\rule{0.22em}{0ex}}\left(t\right)\phantom{\rule{0.22em}{0ex}}and\phantom{\rule{0.22em}{0ex}}with\phantom{\rule{0.22em}{0ex}}x\phantom{\rule{0.22em}{0ex}}.\\ \\ L=side\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}the\phantom{\rule{0.22em}{0ex}}square\phantom{\rule{0.22em}{0ex}}\\ \\ \frac{dB}{dx}=-{10}^{-3}\phantom{\rule{0.22em}{0ex}}T\phantom{\rule{0.22em}{0ex}}c{m}^{-1}=-0.1\phantom{\rule{0.22em}{0ex}}T{m}^{-1}\\ \\ \frac{dB}{dx}=-{10}^{-3}\phantom{\rule{0.22em}{0ex}}T\phantom{\rule{0.22em}{0ex}}se{c}^{-1}\\ \\ \frac{dx}{dt}=8\phantom{\rule{0.22em}{0ex}}cm\phantom{\rule{0.22em}{0ex}}se{c}^{-1}=0.08\phantom{\rule{0.22em}{0ex}}m\phantom{\rule{0.22em}{0ex}}se{c}^{-1}\\ \\ {𝜙}_{B}=B.A\\ \\ \\ Emf\phantom{\rule{0.22em}{0ex}}due\phantom{\rule{0.22em}{0ex}}to\phantom{\rule{0.22em}{0ex}}changing\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}magnetic\phantom{\rule{0.22em}{0ex}}field\phantom{\rule{0.22em}{0ex}}with\phantom{\rule{0.22em}{0ex}}time\phantom{\rule{0.22em}{0ex}}\\ {ℇ}_{1}=-\frac{d\left(BA\right)}{dt}=-A\frac{dB}{dt}=-\left(0.12⨯0.12\right)⨯\left(-{10}^{-3}\right)=0.0144⨯{10}^{-3}\phantom{\rule{0.22em}{0ex}}=1.44⨯{10}^{-5}Volt\\ \end{array}$aMagnetic flux when square is at distance=𝜙B=x+0.12∫xB(0.12)dw 𝜙B=B(0.12)(x+0.12-0.12)=B(0.12)(0.12)=0.0144B d𝜙Bdt=0.0144dBdt=0.0144dBdxdxdt -ℇ2=0.0144⨯(-0.1)⨯(0.08) ℇ2=0.0144⨯(0.1)⨯(0.08)=11.52⨯10-5 Volt Total EMF=ℇ=ℇ1+ℇ2=1.44⨯10-5+11.52⨯10-5=12.96⨯10-5 Volt Current=I=VR=12.96⨯10-54.5⨯10-3=0.0288 A $\begin{array}{l}Magnetic\phantom{\rule{0.22em}{0ex}}flux\phantom{\rule{0.22em}{0ex}}when\phantom{\rule{0.22em}{0ex}}square\phantom{\rule{0.22em}{0ex}}is\phantom{\rule{0.22em}{0ex}}at\phantom{\rule{0.22em}{0ex}}distance={𝜙}_{B}=\underset{x}{\overset{x+0.12}{\int }}B\left(0.12\right)dw\phantom{\rule{0.22em}{0ex}}\\ {𝜙}_{B}=B\left(0.12\right)\left(x+0.12-0.12\right)=B\left(0.12\right)\left(0.12\right)=0.0144B\\ \\ \frac{d{𝜙}_{B}}{dt}=0.0144\frac{dB}{dt}=0.0144\frac{dB}{dx}\frac{dx}{dt}\\ \\ -{ℇ}_{2}=0.0144⨯\left(-0.1\right)⨯\left(0.08\right)\\ \\ {ℇ}_{2}=0.0144⨯\left(0.1\right)⨯\left(0.08\right)=11.52⨯{10}^{-5}\phantom{\rule{0.22em}{0ex}}Volt\\ \\ Total\phantom{\rule{0.22em}{0ex}}EMF=ℇ={ℇ}_{1}+{ℇ}_{2}=1.44⨯{10}^{-5}+11.52⨯{10}^{-5}=12.96⨯{10}^{-5}\phantom{\rule{0.22em}{0ex}}Volt\\ \\ Current=I=\frac{V}{R}=\frac{12.96⨯{10}^{-5}}{4.5⨯{10}^{-3}}=0.0288\phantom{\rule{0.22em}{0ex}}A\\ \phantom{\rule{0.88em}{0ex}}\end{array}$ wx0.12m0.12m